若x属于[-π 6,π 2)

1、f(-π/6)=√2sin(-π/6-π/12)=√2sin(-π/4)=√2*(-√2/2)=-12、因为sinθ=-4/5,θ属于(3π/2,2π)所以cosθ=3/5f(2θ+π/3)=√2

f(x)=2√3sinxcosx+2cosx^2-1=√3sin(2x)+cos(2x)=2sin(2x+π/6）f(x0)=2sin(2x0+π/6)=6/5sin(2x0+π/6)=3/5sin2

（1）F(X)=SIN(X+π/6)+2SIN^2(x/2)=SIN(X+π/6)+1-COSX=SIN(X+π/6)+1-SIN(π/2-X)=2COS[(X+π/6+π/2-X)/2]*SIN[(

f(x)=2cosxcos(x-π/6)-√3sin^2x+sinxcosx=2cosxcos(x-π/6)-√3sin^2x+sinxcosx=2cosx(√3/2cosx+1/2sinx)-√3i

值域：【0.5-b,0.5+b】,b=(根号2)/2,sin2a=1/3+(1/6)*根号(14）f(x)=(1-cos(2x))/2+sin(2x)/2=1/2+c*(sin(2x-pai/4))/

令f(x)=tanx-x,f'(x)=1/cosx^2-1,显然当X属于(0,π/2)时cosx^2＜1所以f'(x)=1/cosx^2-1＞0既f(x)=tanx-x在X属于(0,π/2)时单调递增

tanx>x,这是显然的,利用求导很容易那么等号右边就出来了左边就将x/(1+x*x)-arctanx直接求导,导函数恒小于零,函数递减,x=0时为最大值,所以x/(1+x*x)

fx=2cos^2x+2根号3sinxcosx-1=2cos^2x-1+2根号3sinxcosx根据倍角公式,sin2α=2sinαcosαcos2α=2cos^2（α）-1fx=cos2x+根号3s

1,已知函数f(x)=3sin(2x+π/6),若x属于[-π/6,π/6],求f(x)值域2x+π/6属于[-π/6,π/2]那么sin(2x+π/6)属于【-1/2,1】那么值域是[-3/2,3]

x∈[0,π/2]2x+π/6∈[π/6,7π/6]当2x+π/6=π/2,f(x)有最大值2当2x+π/6=7π/6,f(x)有最小值-1所以值域为[-1,2]

sin(X)函数的最大值是1；X在【0,π/2】x+π/6属于【π/6；2π/3】根据正旋函数的单调性值域属于【1/2,1】!

[-0.5,0.5]

令x-π/8=π得x＝9π/8>2π/3.所以在已知的区间内取不到最小值－1当x＝π/6时,cos(π/6-π/8)＝cosπ/24当x＝2π/3时cos（2π/3-π/8）=cos13π/24cos

f(x)=(1/2)cosx-[(√3)/2]sinx-cosx=-(1/2)cosx-[(√3)/2]sinx=-sin[x+(π/6)]-1≤f(x)≤1

f(x)=cos(x+π/6)-cos(x-π/6)+SQR（3）*cosx=cosxcosπ/6-sinxsinπ/6-cosxcosπ/6-sinxsinπ/6+SQR（3）*cosx=-sinx

sin(π/2+x)=cosx=-4/5x在第二象限所以sinx>0所以sinx=3/5所以原式=cosπ/3cosx+sinπ/3sinx=(-4+3√3)/10

f(x)=2cos(2x-π/3)-√3sin(2x)+1=2*(cos(2x)cos(π/3)+sin(2x)sin(π/3))-√3sin(2x)+1=2*cos(2x)/2+2*sin(2x)*

forx=π/2y=cos(π/2+π/6)=cos(2π/3)=-1/2x属于【0,π/2】的值域是[-1/2,1]

比起跟你讲答案,我更愿意跟你讲思路：对于第一题,可以通过替换将2x+pi/6看成一个变量a,函数的单调区间显然只由sinx函数决定,所以可以令-pi/2+2kpi